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Homework Help: Focal length of thin converging lens

  1. Nov 4, 2015 #1
    1. The problem statement, all variables and given/known data

    An outdoor security camera has a simple lens with focal object length fo = 6.00 cm and perfectly focuses the image of a bird onto its detector located 9.00 cm away from the lens. The animal then flies away from the lens and the new focal distance is 8.00 cm. (a) How far did the bird move? (b) What is the new magnification?

    2. Relevant equations
    \[\frac{1}{f} = \frac{1}{s} + \frac{1}{{s'}}\]
    3. The attempt at a solution

    This is my solution:
    \frac{1}{{{f_0}}} = \frac{1}{s} + \frac{1}{{s'}}\\
    \frac{1}{{{s_1}}} = \frac{1}{{{f}}} - \frac{1}{{s'}} = \frac{1}{6} - \frac{1}{9} = \frac{1}{{18}}\\
    {s_1} = 18cm\\
    \frac{1}{{{s_2}}} = \frac{1}{{{f}}} - \frac{1}{{s'}} = \frac{1}{6} - \frac{1}{8} = \frac{1}{{24}}\\
    {s_2} = 24cm\\
    \Delta {s_o} = {s_2} - {s_1} = 24cm - 18cm = 6cm
    As my textbook offers limited worked solutions for this topic I have looked around and it seems that the focal length does not change on a 'simple lens', even if it is bi-convex. It is s and s prime that changes. Or am I misunderstanding the initial question? The wording is somewhat vague.

    This is the solution I was presented with:

    f1= 6.00 cm

    sā€™1 = sā€™2= 9 cm ( f > sā€™ >2f ) + image is real, inverted, and reduced (for both)

    f2 = 8.00 cm

    s1 = ?

    s2 = ?

    Ds0= s2- s1

    A camera should mean a bi-convex lens.
    \frac{1}{{{f_0}}} = \frac{1}{s} + \frac{1}{{s'}}\\
    \frac{1}{{{s_1}}} = \frac{1}{{{f_1}}} - \frac{1}{{s'}} = \frac{1}{6} - \frac{1}{9} = \frac{1}{{18}}\\
    {s_1} = 18cm\\
    \frac{1}{{{s_2}}} = \frac{1}{{{f_2}}} - \frac{1}{{s'}} = \frac{1}{8} - \frac{1}{9} = \frac{1}{{72}}\\
    {s_2} = 72cm\\
    \Delta {s_o} = {s_2} - {s_1} = 72cm - 18cm = 54cm
    The bird flew 54 cm or just over half a metre from the camera.
  2. jcsd
  3. Nov 4, 2015 #2


    User Avatar
    2017 Award

    Staff: Mentor

    For s2, I think the 8 centimeters are supposed to change to 9 cm (so the distance between detector and lens changes). The lens itself does not change.

    Focal length and focal distance are different things.
  4. Nov 4, 2015 #3
    Thanks for the reply mfb.

    I think I am also confusing focal length and focal point, but the solution I was given still doesn't make sense to me. If the image remains focused on the detector then the lens must change or the detector-lens distance must change. If the detector-lens distance changes then it would be from 9cm to 8cm because the object distance has definitely changed in the wording of the question.
  5. Nov 5, 2015 #4

    rude man

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    Homework Helper
    Gold Member

    Well, you have a good point. A 'simple lens' could not change its focal length.
    However, in this problem it must, in order to keep a sharp image for different object distances s1 and s2.
    So your 'simple' lens has to be a 'zoom' lens. And I think the problem statement is wrong to call it a 'simple lens'.
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